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Hey guys, I solved for the initial velocity which is 24.01m/s. However, when calculating for the final velocity, do you take displacement as -2.48m. Furthermore, when calculating the final velocity why is it that you use the same initial velocity when the particle is moving downwards since acceleration is now 9.8m/s^2 not -9.8. Thanks for your help
(edited 2 years ago)

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Reply 1
Original post by Nithu05
Hey guys, I solved for the initial velocity which is 24.01m/s. However, when calculating for the final velocity, do you take displacement as -2.48m. Furthermore, when calculating the final velocity why is it that you use the same initial velocity when the particle is moving downwards since acceleration is now 9.8m/s^2 not -9.8. Thanks for your help

Yes. If upwards is positive (velocity) then the floor is a downwards displacement of -2.48.
If upwards is positive, the downards acceleration is always -9.8 throughout the flight.
When the ball returns to the initial point, the velocity is the same as the iniital, but negative as the trajectory is a quadratic, symmetric about the highest point.

A sketch with the positive direction clearly marked on usually helps.
(edited 2 years ago)
Reply 2
Original post by mqb2766
Yes. If upwards is positive (velocity) then the floor is a downwards displacement of -2.48.
If upwards is positive, the downards acceleration is always -9.8 throughout the flight.
When the ball returns to the initial point, the velocity is the same as the iniital, but negative as the trajectory is a quadratic, symmetric about the highest point.

Ok so the initial velocity is -24.01m/s when calculating the final velocity? I'm quite sure the mark scheme is wrong. Could you please confirm? Thanks
(edited 2 years ago)
Reply 3
Original post by Nithu05
Ok so the initial velocity is -24.01m/s when calculating the final velocity?

The velocity of the particle when it returns to the initial position is -24.01. It is moving downwards, hence the importance of defining the positive direction clearly and using it consistently.
(edited 2 years ago)
Reply 4
Original post by mqb2766
The velocity of the particle when it returns to the initial position is -24.01. It is moving downwards, hence the importance of defining the positive direction clearly and using it consistently.

Ok thank you so much for your help :redface: Could you take a look at the mark scheme I attached in my previous comment please. Thanks
Reply 5
Original post by Nithu05
Ok thank you so much for your help :redface: Could you take a look at the mark scheme I attached in my previous comment please. Thanks

Yes, what are you unsure about?
They flip the positive direction in the second half of the calculation.
(edited 2 years ago)
Reply 6
Original post by mqb2766
Yes, what are you unsure about?

I'm pretty sure it's wrong since it uses displacement as 2.48 m instead of -2.48m
Reply 7
Original post by Nithu05
I'm pretty sure it's wrong since it uses displacement as 2.48 m instead of -2.48m

Why not do it your way and see if you can work out what they've done / if its correct? What are the signsl of the terms in
v^2 = u^2 + 2as
... does it change the calculation of v?
Reply 8
Original post by mqb2766
Why not do it your way and see if you can work out what they've done / if its correct? What are the signsl of the terms in
v^2 = u^2 + 2as
... does it change the calculation of v?

Yes it does since I'm getting v= 23.49842761 m/s whereas they are getting 25..m/s. I think they may have made an error since their initial velocity and displacement values are both positive when they should be negative
Although the initial velocity value wouldn't make much of a difference when you plug it into the equation.
(edited 2 years ago)
Reply 9
Original post by Nithu05
Yes it does since I'm getting v= 23.49842761 m/s whereas they are getting 25..m/s. I think they may have made an error since their initial velocity and displacement values are both positive when they should be negative

Firstly, if a ball falls a further 2m, the speed must increase. So 23.5 isnt correct.
Write down each of the variables properly and with reference to the positive direction (upwards), like they have in the model solution on the left hand side.
(edited 2 years ago)
Reply 10
Original post by mqb2766
Firstly, if a ball falls a further 2m, the speed must increase. So 23.5 isnt correct.
Write down each of the variables properly and with reference to the positive direction (upwards), like they have in the model solution on the left hand side.

Yes you are correct but I don't know where I went wrong. Could you please check my values?
S=-2.48m U=-24.01m/s V=? A=9.8m/s^2 T=?
Reply 11
Original post by Nithu05
Yes you are correct but I don't know where I went wrong. Could you please check my values?
S=-2.48m U=-24.01m/s V=? A=9.8m/s^2 T=?

Oh no. I see where I went wrong. I didn't multiply it by 2 in the v^2=u^2+2as equation
Original post by Nithu05
Yes you are correct but I don't know where I went wrong. Could you please check my values?
S=-2.48m U=-24.01m/s V=? A=9.8m/s^2 T=?

Which direction does gravity (acceleration) act and what is its sign? Why is this important in the suvat equation? How does it therefore match with the model solution?

Hint A positive acceleration gives a "u" quadratic which will accelerate out into space.
Original post by Nithu05
Oh no. I see where I went wrong. I didn't multiply it by 2 in the v^2=u^2+2as equation

A small error but not the most important. See the previous post.
Reply 14
Original post by mqb2766
Which direction does gravity (acceleration) act and what is its sign? Why is this important in the suvat equation? How does it therefore match with the model solution?

Hint A positive acceleration gives a "u" quadratic which will accelerate out into space.

Gravity normally acts in the downwards direction and is positive. In this case, the particle is moving downwards with an initial velocity of -24.01m/s so shouldn't acceleration be 9.8m/s^2?
Original post by Nithu05
Gravity normally acts in the downwards direction and is positive. In this case, the particle is moving downwards with an initial velocity of -24.01m/s so shouldn't acceleration be 9.8m/s^2?

No. The sign of the acceleration depends on the direction you define as positive, just like displacement and velocity. If you define upwards as positive, a=-9.8. If you define downwards as positive, a=9.8. In this case its the former. In the mark scheme, they flip the positive direction in the second part of the calculation, hence its the latter.
(edited 2 years ago)
Reply 16
Original post by mqb2766
No. The sign of the acceleration depends on the direction you define as positive, just like displacement and velocity. If you define upwards as positive, a=-9.8. If you define downwards as positive, a=9.8. In this case its the former. In the mark scheme, they flip the positive direction in the second part of the calculation, hence its the latter.

I'm slightly confused as to what you mean by 'flip the positive direction'. My initial velocity is negative but isn't the ball travelling downwards so Im confused as to how g is -9.8. Sorry if I am taking up your time
Reply 17
Original post by Nithu05
Hey guys, I solved for the initial velocity which is 24.01m/s. However, when calculating for the final velocity, do you take displacement as -2.48m. Furthermore, when calculating the final velocity why is it that you use the same initial velocity when the particle is moving downwards since acceleration is now 9.8m/s^2 not -9.8. Thanks for your help

Just use conservation of energy: height above the ground is 2.48+2.45*24.01/2. Then use v=sqrt(2gh)
Reply 18
Original post by Dscasg
Just use conservation of energy: height above the ground is 2.48+2.45*24.01/2. Then use v=sqrt(2gh)

Sorry I don't do physics. I'm currently doing mechanics
Original post by Nithu05
I'm slightly confused as to what you mean by 'flip the positive direction'. My initial velocity is negative but isn't the ball travelling downwards so Im confused as to how g is -9.8. Sorry if I am taking up your time

Its worth you working through the calculation properly first (hope you have) and use mark schemes like this only after you are fairly sure about stuff. Otherwise its easy to miss assumptions... like this which maybe they're not clear about.

Firstly, the directlon you assume is positive does not change the calculation or motion. But you have to use it consistently as it determines the signs of the displacement, velocity and acceleration.

If upwards is positive, then the initial velocity is positive (t=0) and the final velocity when it hits the floor is negative. The floor displacement is negative -2.5 and the acceleration is -9.8 because the acceleration acts downards.

In the 2nd half of the model solution, they change positive to act in the downards direction. Soi the initial velocity is -24 (t=0) and when it returns to the initial positon its 24, the displacement is 2.5 and the acceleration is 9.8. They do this just simply to flip all the negative terms to be positive and hopefullly reduce errors with reasoning about negative signs. However, it has to be done clearly.

Both give the same results as the
2as
multiplies two terms a and s, which have the same sign (for this question part). Either both positive or both negative, depending on which direction is positive. However, it does not matter as their product is positive so the speed increases as it falls the further distance to the floor. As mentioned earlier, a clear sketch with the positive direction and the terms marked on helps.
(edited 2 years ago)

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